Respuesta :
Answer:
a) [tex]A_{s}=5.02m^{2}[/tex]
b) [tex]A_{s}=4.75m^{2}[/tex]
c) [tex]A_{s}=4.87m^{2}[/tex]
Explanation:
Water in: [tex]m_{w}=2.5kg/s, C_{w}=4.2kJ/kg\°C, T_{w,in}=15\°C, T_{w,out}=?\°C[/tex]
Process fluid in: [tex]m_{pf}=2kg/s, C_{pf}=3.5kJ/kg\°C, T_{pf,in}=80\°C, T_{pf,out}=50\°C[/tex]
[tex]U=2000W/m^{2}K, LMTD=?\°C, A_{s}=?m^{2}[/tex]
Step 1: Determine the rate of heat transfer in the heat exchanger
[tex]Q=m_{pf}C_{pf}(T_{pf,in}-T_{pf,out})[/tex]
[tex]Q=2*3.5*(80-50)[/tex]
[tex]Q=420kW[/tex]
Step 2: Determine outlet temperature of water
[tex]Q=m_{w}C_{w}(T_{w,in}-T_{w,out})[/tex]
[tex]420=2.5*4.2*(15-T_{w,out})[/tex]
[tex]T_{w,out}=25\°C[/tex]
Part a) Parallel Flow
Step 3: Determine the Logarithmic Mean Temperature Difference (LMTD)
[tex]dT_{1}=T_{pf,in}-T_{w,in}[/tex]
[tex]dT_{1}=80-15[/tex]
[tex]dT_{1}=65\°C[/tex]
[tex]dT_{2}=T_{pf,out}-T_{w,out}[/tex]
[tex]dT_{2}=50-25[/tex]
[tex]dT_{2}=25\°C[/tex]
[tex]LMTD = \frac{dT_{1}-dT_{2}}{ln(\frac{dT_{1}}{dT_{2}})}[/tex]
[tex]LMTD = \frac{65-25}{ln(\frac{65}{25})}[/tex]
[tex]LMTD = \frac{40}{ln(2.6)}[/tex]
[tex]LMTD = 41.86\°C[/tex]
Step 4: Determine required surface area of heat exchanger
[tex]Q=UA_{s}LMTD[/tex]
[tex]420*10^{3}=2000*A_{s}*41.86[/tex]
[tex]A_{s}=5.02m^{2}[/tex]
Part b) Counter Flow
Step 3: Determine the Logarithmic Mean Temperature Difference (LMTD)
[tex]dT_{1}=T_{pf,in}-T_{w,out}[/tex]
[tex]dT_{1}=80-25[/tex]
[tex]dT_{1}=55\°C[/tex]
[tex]dT_{2}=T_{pf,out}-T_{w,in}[/tex]
[tex]dT_{2}=50-15[/tex]
[tex]dT_{2}=35\°C[/tex]
[tex]LMTD = \frac{dT_{1}-dT_{2}}{ln(\frac{dT_{1}}{dT_{2}})}[/tex]
[tex]LMTD = \frac{55-35}{ln(\frac{55}{35})}[/tex]
[tex]LMTD = \frac{20}{ln(1.57)}[/tex]
[tex]LMTD = 44.25\°C[/tex]
Step 4: Determine required surface area of heat exchanger
[tex]Q=UA_{s}LMTD[/tex]
[tex]420*10^{3}=2000*A_{s}*44.25[/tex]
[tex]A_{s}=4.75m^{2}[/tex]
Part c) 1-2 shell and tube exchanger with the water on the shell side
Step 3: Determine the Logarithmic Mean Temperature Difference (LMTD) for counter flow
[tex]dT_{1}=T_{pf,in}-T_{w,out}[/tex]
[tex]dT_{1}=80-25[/tex]
[tex]dT_{1}=55\°C[/tex]
[tex]dT_{2}=T_{pf,out}-T_{w,in}[/tex]
[tex]dT_{2}=50-15[/tex]
[tex]dT_{2}=35\°C[/tex]
[tex]LMTD = \frac{dT_{1}-dT_{2}}{ln(\frac{dT_{1}}{dT_{2}})}[/tex]
[tex]LMTD = \frac{55-35}{ln(\frac{55}{35})}[/tex]
[tex]LMTD = \frac{20}{ln(1.57)}[/tex]
[tex]LMTD = 44.25\°C[/tex]
Step 4: Determine LMTD Correction Factor and Corrected LMTD
[tex]P=\frac{T_{pf,out}-T_{pf,in}}{T_{w,in}-T_{pf,in}}[/tex]
[tex]P=\frac{50-80}{15-80}[/tex]
[tex]P=\frac{-30}{-65}[/tex]
[tex]P=0.462[/tex]
[tex]R=\frac{T_{w,in}-T_{w,out}}{T_{pf,out}-T_{pf,in}}[/tex]
[tex]R=\frac{15-25}{50-80}[/tex]
[tex]R=\frac{-10}{-30}[/tex]
[tex]R=0.333[/tex]
Correction factor F can be determined from attached graph using P and F values calculated above
[tex]F=0.975[/tex]
[tex]LMTD_{corrected}=F*LMTD[/tex]
[tex]LMTD_{corrected}=0.975*44.25[/tex]
[tex]LMTD_{corrected}=43.14\°C[/tex]
Step 5: Determine required surface area of heat exchanger
[tex]Q=UA_{s}LMTD_{corrected}[/tex]
[tex]420*10^{3}=2000*A_{s}*43.14[/tex]
[tex]A_{s}=4.87m^{2}[/tex]
